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Exercise 5.2 · Q8

Q.The AM of two numbers exceeds their GM by 1010 and HM by 1616. Find the numbers.

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Use the identity AM×HM=GM2AM\times HM=GM^2 together with the two given differences to solve for the three means, then recover the two original numbers from their sum and product.

Step 1. Set up the two given conditions. A−G=10A-G=10 and A−H=16A-H=16, where A=AM, G=GM, H=HMA=AM,\ G=GM,\ H=HM.

Step 2. Express G,HG,H in terms of AA. G=A−10G=A-10, H=A−16H=A-16.

Step 3. Use AM×HM=GM2AM\times HM=GM^2 (Result 5.3). A⋅H=G2⇒A(A−16)=(A−10)2A\cdot H = G^2 \Rightarrow A(A-16)=(A-10)^2.

Step 4. Expand and solve for AA.

A2−16A=A2−20A+100⇒−16A+20A=100⇒4A=100⇒A=25A^2-16A = A^2-20A+100 \Rightarrow -16A+20A=100 \Rightarrow 4A=100 \Rightarrow A=25.

Step 5. Recover G,HG,H. G=25−10=15G=25-10=15,  H=25−16=9\ H=25-16=9. …

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